Two Threes, Read Against Each Other
The first argument on this ladder that reads two numbers at once. Two of them, both worth deriving rather than learning by heart, and both available on a blank board.
Side by side
Two threes sharing a side settle three parallel segments and two crossings. The far side of each one is drawn, the side between them is drawn, and the two segments continuing that middle side beyond the pair are crossed out. Five segments from two numbers, on a board where nothing has happened yet.
The far sides are the part you can prove in your head. Suppose the left three’s far side were crossed: it would need its other three, so its top, its bottom and the shared side would all be drawn. But the dot where its top meets the shared side now has two lines, which crosses out the right three’s top — and the dot at the other end of the shared side does the same to the right three’s bottom. That leaves the right three with a single side available for a number that wants three, which is impossible, so the far side was drawn all along. The same argument mirrored gives the other far side.
The middle segment is a different kind of claim and worth being honest about. Crossing it out breaks no number and no dot; what it does is force the eight segments around the pair into a closed ring of their own. That ring is a perfectly legal loop — it is just not this board’s loop, because the numbers everywhere else would then all have to be zeros. So the middle segment is drawn, and the reason is the closing rule two rungs up, borrowed early because the pattern is worth recognising on sight.
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Corner to corner
Two threes touching only at a corner settle four segments: on each of them, the two sides facing away from the shared corner are drawn. Nothing else about either square is decided, and nothing about the dot they share.
The proof is shorter than the one above. Suppose the top side of the upper three were crossed out. It would then need its other three sides, and two of those meet at the shared corner dot — which is now full. Both of the lower three’s sides at that dot are therefore crossed out, leaving it two sides for a number that wants three. Contradiction, so the top side is drawn, and the same argument covers the other three cases.
This one comes up more often than the side-by-side version on the boards here, largely because diagonal contact is simply more common than shared sides on a wandering loop. It is also the more useful of the two on a large board: four drawn segments in two separate places give the dot rule something to work with in two separate regions at once, and a board that had nothing to say five seconds earlier starts moving in both.
What a hard board actually contains
A hard board here is one where the solver, always taking the cheapest available move, was at some point forced to read two threes together — no zero left unused, no dot with a line and one way out, no number that could be counted against its own sides, and no corner left to read. That is a strong statement about the board rather than about its size: it contains a genuine bottleneck.
It is also why size and difficulty are separate choices on this site. A 12×12 easy board has a hundred and forty-four squares and no bottleneck at all; a 6×6 hard board has thirty-six and one. Most players find the second takes longer, and almost everybody finds it more satisfying.
Where this sits on the ladder
- Rung 1Counting a cell
A number with its sides already drawn has none left to give; a zero gives none at all.
- Rung 2Following the line
Every dot carries no line or exactly two, so a line that arrives has to leave again.
- Rung 3The corner cases
A corner dot has only two segments to offer, which settles a 1, a 2 or a 3 sitting in it.
- Rung 4Threes together
Two threes side by side, or corner to corner, decide their outer sides before anything is drawn.
- Rung 5No early closing
A segment that would seal a short ring with numbers still unsatisfied can never be drawn.
- Rung 6Assumption
Draw a segment, follow it until the board breaks, and rub it out again.
Common questions
Why are two adjacent threes always joined by a line?
Because crossing the side between them forces the eight segments around the pair into a closed ring of their own, and that ring can only be the answer if every other number on the board is a zero.
What do two diagonal threes give you?
Four drawn segments: on each three, the two sides facing away from the corner they share. Crossing any of them would fill the shared dot and leave the other three a side short.
Do these patterns work with other numbers?
Not in this form. They rely on a 3 having only one spare side, which is what makes a single crossed-out side fatal. Adjacent 2s and adjacent 1s say nothing on their own.
More Slitherlink pages
- SlitherlinkUnlimited boards
- Daily slitherlinkA new loop every day
- Slitherlink rulesThree rules, in full
- Solving a loopA 6×6, worked
- Loop techniquesThe whole ladder
- Squares and dotsTechnique — easy
- The corner casesTechnique — medium
- Closing too earlyTechnique — expert
- Guessing at loopsTechnique — refused, here too
- Printable slitherlinkFor paper
- Slitherlink archiveEvery past loop